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%I #17 Dec 17 2018 06:00:13
%S 3,10,4,7,10,2,10,6,5,6,5,10,7,10,8,10,2,10,10,2,10,7,4,10,10,8,3,4,
%T 10,3,3,3,10,10,4,10,3,10,10,6,7,10
%N Minimal k > 1 such that the base-k representation of the n-th Ramanujan prime (A104272), read in decimal, is also a Ramanujan prime.
%C Conjecture 1. Every number 2,...,10 occurs infinitely many times.
%C Conjecture 2. There exists limit of average (a(1)+...+a(n))/n.
%C However note that, already for the eleventh Ramanujan prime 101 the binary representation, read in decimal, is rather large prime: 1100101. We cannot decide, if it is a Ramanujan prime, using b-file in A104272. Therefore, the calculation of the sequence here ends off. It is a general problem: how to decide, if a large prime is a Ramanujan one?
%e The third Ramanujan prime is 17. If k=2, we have 10001, if k=3, we have 122, if k=4, we have 101. In this list, read in decimal, 101 is the first prime. Since 101 is a Ramanujan prime, then a(3)=4.
%o (PARI) ramanujan_prime_list(n) = {my(L=vector(n), s=0, k=1); for(k=1, prime(3*n)-1, if(isprime(k), s++); if(k%2==0 && isprime(k/2), s--); if(s<n, L[s+1] = k+1)); L;} \\ A104272
%o a(n, vp, pmax) = {my(p=vp[n], d, np); for (b=2, 10, d = digits(p, b); np = fromdigits(d, 10); if (np > pmax, return (0)); if (vecsearch(vp, np), return (b)););}
%o lista(nn) = {my(vp = ramanujan_prime_list(nn), pmax = vecmax(vp)); for (n=1, nn, my(result = a(n, vp, pmax)); if (result, print1(result, ", "), break););} \\ use nn=10^7 to get 42 terms \\ _Michel Marcus_, Dec 17 2018
%Y Cf. A104272, A235354.
%K nonn,base,more
%O 1,1
%A _Vladimir Shevelev_, Jan 17 2014
%E a(11)-a(42) from _Michel Marcus_, Dec 17 2018
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